从列表中删除第一项的最快方法是什么？

Question

我使用list阅读了一些快速as a reference操作，正如here所述。

从列表中删除第一项的最快方法是什么（可能使用as a reference和rest of list）？

E.g：

{3,5,6,2,8}

变成

{5,6,2,8}

Answer 1

这将删除列表中的第一项

set theList to rest of {3, 5, 6, 2, 8}

返回{5,6,2,8}

或者，这将删除列表中的最后一项

set theList to reverse of rest of reverse of {3, 5, 6, 2, 8}

返回{3,5,6,2}

Answer 2

在您的示例列表中，{3, 5, 6, 2, 8}，请将其设置为：

set theList to {3, 5, 6, 2, 8}

然后从列表中删除3的一种方法是：

set theList to items 2 thru -1 of theList

或者：

set theList to items 2 thru -1 of {3, 5, 6, 2, 8}

Answer 3

--use my handler 

set b to {"A", 2, 3}

removeAnItemInList(1, b)
--result: {2, 3}



to removeAnItemInList(DeleteOffset, Listt)
    
    ---TH là 1 thì biến 1 thành {}
    if (count Listt) is 1 and (DeleteOffset is 1) then
        set Listt to {}
        return Listt
    end if
    
    --- TH  là 2 thì bỏ 1 hoặc 2
    if (count Listt) is 2 and (DeleteOffset is 1) then
        set Listt to item 2 of Listt as list
        return Listt
    end if
    
    if (count Listt) is 2 and (DeleteOffset is 2) then
        set Listt to item 1 of Listt as list
        return Listt
    end if
    
    ---TH >2 bỏ đầu cuối ----
    if (count Listt) > 2 and (DeleteOffset is 1) then
        set Listt to items 2 thru end of Listt
        return Listt
    end if
    
    if (count Listt) > 2 and (DeleteOffset is (count Listt)) then
        set Listt to items 1 thru -2 of Listt
        return Listt
    end if
    
    ---TH >2 bỏ giữa
    if (count Listt) > 2 then
        set Listt to items 1 thru (DeleteOffset - 1) of Listt & items (DeleteOffset + 1) thru end of Listt
        return Listt
    end if
    
end removeAnItemInList